3.2.77 \(\int \frac {x^{9/2} (A+B x)}{(b x+c x^2)^2} \, dx\)

Optimal. Leaf size=131 \[ -\frac {b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}}+\frac {b \sqrt {x} (7 b B-5 A c)}{c^4}-\frac {x^{3/2} (7 b B-5 A c)}{3 c^3}+\frac {x^{5/2} (7 b B-5 A c)}{5 b c^2}-\frac {x^{7/2} (b B-A c)}{b c (b+c x)} \]

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Rubi [A]  time = 0.07, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {781, 78, 50, 63, 205} \begin {gather*} -\frac {b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}}+\frac {x^{5/2} (7 b B-5 A c)}{5 b c^2}-\frac {x^{3/2} (7 b B-5 A c)}{3 c^3}+\frac {b \sqrt {x} (7 b B-5 A c)}{c^4}-\frac {x^{7/2} (b B-A c)}{b c (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^2,x]

[Out]

(b*(7*b*B - 5*A*c)*Sqrt[x])/c^4 - ((7*b*B - 5*A*c)*x^(3/2))/(3*c^3) + ((7*b*B - 5*A*c)*x^(5/2))/(5*b*c^2) - ((
b*B - A*c)*x^(7/2))/(b*c*(b + c*x)) - (b^(3/2)*(7*b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(9/2)

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 781

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^{9/2} (A+B x)}{\left (b x+c x^2\right )^2} \, dx &=\int \frac {x^{5/2} (A+B x)}{(b+c x)^2} \, dx\\ &=-\frac {(b B-A c) x^{7/2}}{b c (b+c x)}-\frac {\left (-\frac {7 b B}{2}+\frac {5 A c}{2}\right ) \int \frac {x^{5/2}}{b+c x} \, dx}{b c}\\ &=\frac {(7 b B-5 A c) x^{5/2}}{5 b c^2}-\frac {(b B-A c) x^{7/2}}{b c (b+c x)}-\frac {(7 b B-5 A c) \int \frac {x^{3/2}}{b+c x} \, dx}{2 c^2}\\ &=-\frac {(7 b B-5 A c) x^{3/2}}{3 c^3}+\frac {(7 b B-5 A c) x^{5/2}}{5 b c^2}-\frac {(b B-A c) x^{7/2}}{b c (b+c x)}+\frac {(b (7 b B-5 A c)) \int \frac {\sqrt {x}}{b+c x} \, dx}{2 c^3}\\ &=\frac {b (7 b B-5 A c) \sqrt {x}}{c^4}-\frac {(7 b B-5 A c) x^{3/2}}{3 c^3}+\frac {(7 b B-5 A c) x^{5/2}}{5 b c^2}-\frac {(b B-A c) x^{7/2}}{b c (b+c x)}-\frac {\left (b^2 (7 b B-5 A c)\right ) \int \frac {1}{\sqrt {x} (b+c x)} \, dx}{2 c^4}\\ &=\frac {b (7 b B-5 A c) \sqrt {x}}{c^4}-\frac {(7 b B-5 A c) x^{3/2}}{3 c^3}+\frac {(7 b B-5 A c) x^{5/2}}{5 b c^2}-\frac {(b B-A c) x^{7/2}}{b c (b+c x)}-\frac {\left (b^2 (7 b B-5 A c)\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^2} \, dx,x,\sqrt {x}\right )}{c^4}\\ &=\frac {b (7 b B-5 A c) \sqrt {x}}{c^4}-\frac {(7 b B-5 A c) x^{3/2}}{3 c^3}+\frac {(7 b B-5 A c) x^{5/2}}{5 b c^2}-\frac {(b B-A c) x^{7/2}}{b c (b+c x)}-\frac {b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 110, normalized size = 0.84 \begin {gather*} \frac {\sqrt {x} \left (b^2 (70 B c x-75 A c)-2 b c^2 x (25 A+7 B x)+2 c^3 x^2 (5 A+3 B x)+105 b^3 B\right )}{15 c^4 (b+c x)}-\frac {b^{3/2} (7 b B-5 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^2,x]

[Out]

(Sqrt[x]*(105*b^3*B + 2*c^3*x^2*(5*A + 3*B*x) - 2*b*c^2*x*(25*A + 7*B*x) + b^2*(-75*A*c + 70*B*c*x)))/(15*c^4*
(b + c*x)) - (b^(3/2)*(7*b*B - 5*A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(9/2)

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IntegrateAlgebraic [A]  time = 0.14, size = 119, normalized size = 0.91 \begin {gather*} \frac {\left (5 A b^{3/2} c-7 b^{5/2} B\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{9/2}}+\frac {\sqrt {x} \left (-75 A b^2 c-50 A b c^2 x+10 A c^3 x^2+105 b^3 B+70 b^2 B c x-14 b B c^2 x^2+6 B c^3 x^3\right )}{15 c^4 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(9/2)*(A + B*x))/(b*x + c*x^2)^2,x]

[Out]

(Sqrt[x]*(105*b^3*B - 75*A*b^2*c + 70*b^2*B*c*x - 50*A*b*c^2*x - 14*b*B*c^2*x^2 + 10*A*c^3*x^2 + 6*B*c^3*x^3))
/(15*c^4*(b + c*x)) + ((-7*b^(5/2)*B + 5*A*b^(3/2)*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(9/2)

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fricas [A]  time = 0.42, size = 290, normalized size = 2.21 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{3} - 5 \, A b^{2} c + {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x + 2 \, c \sqrt {x} \sqrt {-\frac {b}{c}} - b}{c x + b}\right ) - 2 \, {\left (6 \, B c^{3} x^{3} + 105 \, B b^{3} - 75 \, A b^{2} c - 2 \, {\left (7 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt {x}}{30 \, {\left (c^{5} x + b c^{4}\right )}}, -\frac {15 \, {\left (7 \, B b^{3} - 5 \, A b^{2} c + {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) - {\left (6 \, B c^{3} x^{3} + 105 \, B b^{3} - 75 \, A b^{2} c - 2 \, {\left (7 \, B b c^{2} - 5 \, A c^{3}\right )} x^{2} + 10 \, {\left (7 \, B b^{2} c - 5 \, A b c^{2}\right )} x\right )} \sqrt {x}}{15 \, {\left (c^{5} x + b c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

[-1/30*(15*(7*B*b^3 - 5*A*b^2*c + (7*B*b^2*c - 5*A*b*c^2)*x)*sqrt(-b/c)*log((c*x + 2*c*sqrt(x)*sqrt(-b/c) - b)
/(c*x + b)) - 2*(6*B*c^3*x^3 + 105*B*b^3 - 75*A*b^2*c - 2*(7*B*b*c^2 - 5*A*c^3)*x^2 + 10*(7*B*b^2*c - 5*A*b*c^
2)*x)*sqrt(x))/(c^5*x + b*c^4), -1/15*(15*(7*B*b^3 - 5*A*b^2*c + (7*B*b^2*c - 5*A*b*c^2)*x)*sqrt(b/c)*arctan(c
*sqrt(x)*sqrt(b/c)/b) - (6*B*c^3*x^3 + 105*B*b^3 - 75*A*b^2*c - 2*(7*B*b*c^2 - 5*A*c^3)*x^2 + 10*(7*B*b^2*c -
5*A*b*c^2)*x)*sqrt(x))/(c^5*x + b*c^4)]

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giac [A]  time = 0.16, size = 122, normalized size = 0.93 \begin {gather*} -\frac {{\left (7 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {B b^{3} \sqrt {x} - A b^{2} c \sqrt {x}}{{\left (c x + b\right )} c^{4}} + \frac {2 \, {\left (3 \, B c^{8} x^{\frac {5}{2}} - 10 \, B b c^{7} x^{\frac {3}{2}} + 5 \, A c^{8} x^{\frac {3}{2}} + 45 \, B b^{2} c^{6} \sqrt {x} - 30 \, A b c^{7} \sqrt {x}\right )}}{15 \, c^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

-(7*B*b^3 - 5*A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + (B*b^3*sqrt(x) - A*b^2*c*sqrt(x))/((c*x +
b)*c^4) + 2/15*(3*B*c^8*x^(5/2) - 10*B*b*c^7*x^(3/2) + 5*A*c^8*x^(3/2) + 45*B*b^2*c^6*sqrt(x) - 30*A*b*c^7*sq
rt(x))/c^10

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maple [A]  time = 0.07, size = 139, normalized size = 1.06 \begin {gather*} \frac {2 B \,x^{\frac {5}{2}}}{5 c^{2}}+\frac {5 A \,b^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{3}}-\frac {7 B \,b^{3} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{4}}-\frac {A \,b^{2} \sqrt {x}}{\left (c x +b \right ) c^{3}}+\frac {2 A \,x^{\frac {3}{2}}}{3 c^{2}}+\frac {B \,b^{3} \sqrt {x}}{\left (c x +b \right ) c^{4}}-\frac {4 B b \,x^{\frac {3}{2}}}{3 c^{3}}-\frac {4 A b \sqrt {x}}{c^{3}}+\frac {6 B \,b^{2} \sqrt {x}}{c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(9/2)*(B*x+A)/(c*x^2+b*x)^2,x)

[Out]

2/5/c^2*B*x^(5/2)+2/3/c^2*A*x^(3/2)-4/3/c^3*B*x^(3/2)*b-4/c^3*A*b*x^(1/2)+6/c^4*b^2*B*x^(1/2)-b^2/c^3*x^(1/2)/
(c*x+b)*A+b^3/c^4*x^(1/2)/(c*x+b)*B+5*b^2/c^3/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*A-7*b^3/c^4/(b*c)^(1
/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*B

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maxima [A]  time = 1.22, size = 115, normalized size = 0.88 \begin {gather*} \frac {{\left (B b^{3} - A b^{2} c\right )} \sqrt {x}}{c^{5} x + b c^{4}} - \frac {{\left (7 \, B b^{3} - 5 \, A b^{2} c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {2 \, {\left (3 \, B c^{2} x^{\frac {5}{2}} - 5 \, {\left (2 \, B b c - A c^{2}\right )} x^{\frac {3}{2}} + 15 \, {\left (3 \, B b^{2} - 2 \, A b c\right )} \sqrt {x}\right )}}{15 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(9/2)*(B*x+A)/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

(B*b^3 - A*b^2*c)*sqrt(x)/(c^5*x + b*c^4) - (7*B*b^3 - 5*A*b^2*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4)
+ 2/15*(3*B*c^2*x^(5/2) - 5*(2*B*b*c - A*c^2)*x^(3/2) + 15*(3*B*b^2 - 2*A*b*c)*sqrt(x))/c^4

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mupad [B]  time = 1.04, size = 146, normalized size = 1.11 \begin {gather*} x^{3/2}\,\left (\frac {2\,A}{3\,c^2}-\frac {4\,B\,b}{3\,c^3}\right )-\sqrt {x}\,\left (\frac {2\,b\,\left (\frac {2\,A}{c^2}-\frac {4\,B\,b}{c^3}\right )}{c}+\frac {2\,B\,b^2}{c^4}\right )+\frac {2\,B\,x^{5/2}}{5\,c^2}+\frac {\sqrt {x}\,\left (B\,b^3-A\,b^2\,c\right )}{x\,c^5+b\,c^4}-\frac {b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,\sqrt {x}\,\left (5\,A\,c-7\,B\,b\right )}{7\,B\,b^3-5\,A\,b^2\,c}\right )\,\left (5\,A\,c-7\,B\,b\right )}{c^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(9/2)*(A + B*x))/(b*x + c*x^2)^2,x)

[Out]

x^(3/2)*((2*A)/(3*c^2) - (4*B*b)/(3*c^3)) - x^(1/2)*((2*b*((2*A)/c^2 - (4*B*b)/c^3))/c + (2*B*b^2)/c^4) + (2*B
*x^(5/2))/(5*c^2) + (x^(1/2)*(B*b^3 - A*b^2*c))/(b*c^4 + c^5*x) - (b^(3/2)*atan((b^(3/2)*c^(1/2)*x^(1/2)*(5*A*
c - 7*B*b))/(7*B*b^3 - 5*A*b^2*c))*(5*A*c - 7*B*b))/c^(9/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(9/2)*(B*x+A)/(c*x**2+b*x)**2,x)

[Out]

Timed out

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